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3's & 7's (Posted on 2006-04-09) Difficulty: 3 of 5
Find the smallest number comprised of only 3’s and 7’s which fits the following conditions:

1) It has at least one 3;
2) It has at least one 7;
3) It is divisible by 3;
4) It is divisible by 7;
5) The sum of its digits is divisible by 3;
6) The sum of its digits is divisible by 7.

See The Solution Submitted by Jer    
Rating: 3.2500 (4 votes)

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Solution Solution (with some trial and error) | Comment 1 of 13
  Conditions 1 and 2 are given. Conditions 3 and 5 are equivalent since the sum of the digits of all numbers divisible by 3 is also divisible by 3. Condition 6 requires the quantity of 3's to be divisible by 7. Condition 5 requires the quantity of 7's to be divisible by 3. Working to satisfy Condition 4; starting with seven 3's and three 7's.

3333333777 = 4 mod 7

3333337377 = 6 mod 7

3333337737 = 2 mod 7

3333337773 = 3 mod 7

3333373377 = 5 mod 7

3333373737 = 1 mod 7

3333373773 = 2 mod 7

3333377337 = 3 mod 7

3333377373 = 4 mod 7

3333377733 = 0 mod 7

The sum of the digits is 42 which is divisible by 3 and 7

and 3333377733/3 = 1111125911 and 3333377733/7 = 476196819

Edited to correct the "mod" format.

Edited on April 9, 2006, 9:24 pm
  Posted by Leming on 2006-04-09 18:22:45

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